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Question
The population of a city increases each year by 4% of what it had been at the beginning of each year. If the population in 1999 had been 6760000, find the population of the city in (i) 2001 (ii) 1997.
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Solution
\[(i)\]
Population of the city in 2001 = P \[\left( 1 + \frac{R}{100} \right)^2 \]
\[ = 6760000 \left( 1 + \frac{4}{100} \right)^2 \]
\[ = 6760000 \left( 1 . 04 \right)^2 \]
\[ = 7311616\]
Thus, Population of the city in 2001 is 7311616.
\[(ii)\]
Population of the city in 1997 = P \[\left( 1 + \frac{R}{100} \right)^{- 2} \]
\[ = 6760000 \left( 1 + \frac{4}{100} \right)^{- 2} \]
\[ = 6760000 \left( 1 . 04 \right)^{- 2} \]
\[ = 6250000\]
Thus, Population of the city in 1997 is 6250000.
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